[LeetCode] Two sum (easy)

twosum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.

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給定兩個非空的鍊結串列，分別代表兩個非負的整數串列，串列內每個元素皆由反方向儲存(越前面的節點位數越低)，每一個節點代表一個位元。
將此兩個鍊結串列相加並回傳。

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Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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參考答案：

int* twoSum(int* nums, int numsSize, int target){
    int *ans = malloc(sizeof(int)*2);
    for(int i=0;i<numsSize;i++){
        for(int j=0;j<numsSize;j++){
            if(target == (nums[i] + nums[j]) && i!=j ){
                ans[0] = i,ans[1] = j;
                return ans;
            }
        }
    }
    return ans;
}

學習目的：

學習兩個以上的回傳值型態宣告方法 malloc

c++ define:
void* malloc (size_t size);

//example
int *data = malloc(1000, sizeof(int)); //宣告
free(data); //釋放

測試程式:

//測試程式
int* twoSum(int* nums, int numsSize, int target){
    int *ans = malloc(sizeof(int)*2);
    for(int i=0;i<numsSize;i++){
        for(int j=0;j<numsSize;j++){
            if(target == (nums[i] + nums[j]) && i!=j ){
                ans[0] = i,ans[1] = j;
                return ans;
            }
        }
    }
    return ans;
}


int main(){
    int nums[] = {2, 7, 11, 15};
    int target = 9;

    printf("sizeof(nums) = %1lu \n", sizeof(nums)/sizeof(int));

    int *out = twoSum(nums, sizeof(nums)/sizeof(int), target);

    printf("\n%d\n",out[0]);
    printf("%d\n",out[1]);

    return 0;
}